# Why work is defined as a scalar?

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1. Jun 13, 2015

### Tahmeed

we know work is F.s which is a scalar quantity. but why it is defined this way??? work actually has something to do with direction. Negative work means reduction in velocity and velocity has direction. What was the problem of defining work as vector that lies in the direction of displacement??

2. Jun 13, 2015

### Staff: Mentor

It does not.
It does not have to be a reduction in velocity, it can be everything. Also, you can change the direction of velocity without altering speed. Do you want to have a separate work quantity for every possible direction where deceleration can occur?
It does not make sense because work does not care about the direction of motion of whatever object was used. It's like assigning a temperature a material property based on the material used to measure it. "It is 30 degrees mercury-celsius today" - "oh, according to my thermometer it is 30 degrees semiconductor-celsius".

3. Jun 13, 2015

### Staff: Mentor

The quantity F.s is a useful quantity. Because it is useful, it is used often, and it was therefore convenient to give it a name. There really isn't any reason for a definition other than usefulness and convenience.

You mean something like $|\mathbf{F}|\mathbf{s}$? Where is such a quantity used? I cannot think of a single time I have ever seen that quantity used.

4. Jun 13, 2015

### theodoros.mihos

5. Jun 15, 2015

### stedwards

Let's look at work as force acting over a distance.

First the force vector:

Fortunately it has units of work, not force, and everything makes sense.

$W[ newton \cdot meter ] = F_x[newton] dx[meter] + F_y[newton] dy[meter] + F_z[newton] dz [meter]$

Normally, you are taught that $F = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}$.

i, j and k are unit vectors in the x, y, and z directions, respectively. The little dx, dy, and dz replace these.

6. Jun 15, 2015

### stedwards

Force is a "covector" having units of work. The vector elements have units of force.

$$w = F_x dx + F_y dy + F_z dz$$
Integrated over a spatial interval, $W_{ab} = \int_{a}^{b} w$

7. Jun 15, 2015

### Fredrik

Staff Emeritus
Work is supposed to be a measure of how "hard" it is to do something. If work had been a vector in the direction of the displacement, then if you push a heavy object 5 meters in one direction and then push it back, you will have performed zero work.

Also, what if you push an object in a curved line? What would be the direction of the work vector then?

8. Jun 15, 2015

### Khashishi

Mathematically, work is the dot product between the force and distance. Dot product creates a scalar out of two vectors. Note that if you push the side of a truck that happens to be moving forward (perpendicular to your force), then you aren't doing work on the truck. The dot product converts the two directions (your force and the trucks motion) into a single quantity, independent of direction, since you can rotate the entire coordinate system and you still do no work on the truck.

Work shares the same units as energy, and energy has no direction. In fact, in special relativity, energy and momentum form a 4-vector. You could think of the energy as the component of energy-momentum directed in the time direction, while the momentum is the three components directed in space directions.

9. Jun 15, 2015

### Khashishi

What? That makes no sense. Force has units of mass*length/time^2. Work has units of mass*length^2/time^2.
Note that length^2 = length dot length, which is a scalar.

10. Jun 15, 2015

### stedwards

$W[ML^2T^{-2}] = F_i[MLT^{-2}]dx^i[L]$

$W_{ab} = \int_a^b F_i dx^i = \int_a^b W$

Re. length squared as a scalar. Interesting.

$\frac{\partial}{\partial x} dx$ is scalar and dimensionless.

Last edited: Jun 15, 2015
11. Jun 18, 2015